python - 如何根据条件拆分列表？

Question

从美学和性能的角度来看，根据条件将项目列表拆分为多个列表的最佳方法是什么？相当于:

good = [x for x in mylist if x in goodvals]
bad  = [x for x in mylist if x not in goodvals]

有没有更优雅的方法来做到这一点？

这是实际用例，以更好地解释我想要做的事情:

# files looks like: [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ... ]
IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
images = [f for f in files if f[2].lower() in IMAGE_TYPES]
anims  = [f for f in files if f[2].lower() not in IMAGE_TYPES]

Answer 1

good, bad = [], []
for x in mylist:
    (bad, good)[x in goodvals].append(x)